What is the solution for y^{''}+8y^'+52y=0,y(0)=6,y^'(0)=7 ?

The solution for y^{''}+8y^'+52y=0,y(0)=6,y^'(0)=7 is y=e^{-4t}(6cos(6t)+31/6 sin(6t))

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y^{''}+8y^'+52y=0,y(0)=6,y^'(0)=7

y^{^{''}} + 8 y^{^{'}} + 52 y = 0, y (0) = 6, y^{^{'}} (0) = 7

y = e^{- 4 t} (6 cos (6 t) + \frac{31}{6} sin (6 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 52 y = 0

Solve linear ODE:

y = e^{- 4 t} (6 cos (6 t) + \frac{31}{6} sin (6 t))

y = e^{- 4 t} (6 cos (6 t) + \frac{31}{6} sin (6 t))

\frac{d ^{2} x}{d t ^{2}} = - π^{2} x

u^{^{''}} + 0.25 u^{^{'}} + H (t) = 0

y^{^{'''}} - 23 y^{^{''}} + 160 y^{^{'}} - 300 y = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 7, y (0) = 7

y^{^{''}} - 4 y^{^{'}} + 5 y = 0

What is the solution for y^{''}+8y^'+52y=0,y(0)=6,y^'(0)=7 ?
The solution for y^{''}+8y^'+52y=0,y(0)=6,y^'(0)=7 is y=e^{-4t}(6cos(6t)+31/6 sin(6t))